The combustion of 132 g of propane, C3H8, with excess oxygen liberates 6.6x10^3 kj of heat. What is the enthalpy change for the reaction below:
C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)
To find the enthalpy change for the given reaction, you need to use the molar masses of each compound and apply the concept of stoichiometry.
The balanced equation shows that for every 1 mole of propane (C3H8), we produce 3 moles of carbon dioxide (CO2) and 4 moles of water (H2O).
First, calculate the moles of propane using its molar mass:
Molar Mass of C3H8 = (12.01 g/mol x 3) + (1.01 g/mol x 8) = 44.11 g/mol
Moles of C3H8 = Mass of C3H8 / Molar Mass of C3H8
Moles of C3H8 = 132 g / 44.11 g/mol
Moles of C3H8 = 2.99 mol (approximately)
Next, use the stoichiometry to find the moles of water produced:
Moles of H2O = 4 moles of H2O / 1 mole of C3H8 x 2.99 moles of C3H8
Moles of H2O = 11.96 mol (approximately)
Now, calculate the heat released per mole of propane combusted:
Heat per mole of C3H8 = -6.6 x 10^3 kJ / 2.99 mol
Heat per mole of C3H8 = -2207 kJ/mol (approximately)
Since the reaction produces 3 moles of CO2 and 4 moles of H2O, the enthalpy change for the reaction can be calculated as follows:
Enthalpy change = (3 moles of CO2 x Heat per mole of CO2) + (4 moles of H2O x Heat per mole of H2O)
Enthalpy change = (3 moles of CO2 x 0 kJ/mol) + (11.96 moles of H2O x Heat per mole of H2O)
Enthalpy change = 0 + (11.96 moles x -2207 kJ/mol)
Enthalpy change = -26381.72 kJ/mol (approximately)
Therefore, the enthalpy change for the given reaction is approximately -26381.72 kJ/mol.
To calculate the enthalpy change for the given reaction, you need to determine the amount of heat released per mole of propane combusted.
Step 1: Calculate the moles of propane (C3H8):
To convert grams of propane to moles, you need to know the molar mass of propane (C3H8). The molar mass of carbon (C) is 12.01 g/mol, and hydrogen (H) is 1.01 g/mol. Multiplying the molar mass by the number of atoms in each element, you get:
3 moles of carbon (3 x 12.01 g/mol) = 36.03 g/mol
8 moles of hydrogen (8 x 1.01 g/mol) = 8.08 g/mol
Adding these values, you get 44.11 g/mol as the molar mass of propane.
To find the number of moles, divide the given mass of propane (132 g) by the molar mass:
132 g / 44.11 g/mol = 2.99 mol (approximately)
Step 2: Calculate the heat released per mole of propane:
The given heat of combustion is 6.6x10^3 kJ. Divide this by the moles of propane (2.99 mol) to find the heat released per mole:
6.6x10^3 kJ / 2.99 mol = 2.21x10^3 kJ/mol
Step 3: Determine the enthalpy change:
In the balanced equation, the stoichiometric coefficients represent the molar ratios of the reactants and products. From the balanced equation:
1 mole of C3H8 produces 3 moles of CO2 and 4 moles of H2O
Therefore, the heat released per mole of CO2 is the same as per mole of propane:
2.21x10^3 kJ/mol
Hence, the enthalpy change for the reaction is 2.21x10^3 kJ/mol.